Reworked the order chapter.

src/fa/order/lattice.tex

@@ -1,63 +1,6 @@
 \section{Vector Lattices}
 \label{section:vector-lattice}
 
-\begin{definition}[Ordered Vector Space]
-\label{definition:ordered-vector-space}
-    Let $E$ be a vector space over $\real$ and $\le$ be a partial order on $E$, then $(E, \le)$ is a \textbf{ordered vector space} if
-    \begin{enumerate}
-        \item[(LO1)] For any $x, y, z \in E$ with $x \le y$, $x + z \le y + z$.
-        \item[(LO2)] For any $x, y \in E$ and $\lambda > 0$, $x \le y$ implies that $\lambda x \le \lambda y$.
-    \end{enumerate}
-
-    
-\end{definition}
-
-\begin{proposition}
-\label{proposition:ordered-vector-space-properties}
-    Let $(E, \le)$ be an ordered vector space and $A, B \subset E$ such that $\sup(A)$ and $\sup (B)$ exist, then 
-    \begin{enumerate}
-        \item $\sup(A + B) = \sup(A) + \sup(B)$.
-        \item $\sup(A) = -\inf (-A)$
-    \end{enumerate}
-
-    
-\end{proposition}
-\begin{proof}
-    (1): For any $a \in A$ and $b \in B$, $a + b \le \sup(A) + \sup(B)$ by (LO1). Let $c \in E$ such that $c \ge a + b$ for all $a \in A$ and $b \in B$, then $c \ge a + \sup(B)$ for all $a \in A$, so $c \ge \sup(A) + \sup(B)$. Therefore $\sup(A) + \sup(B) = \sup(A + B)$.
-
-    (2): For any $a \in A$, $\sup(A) \ge a = -(-a)$, so $-\sup(A) \le \inf(-A)$, and $\sup(A) \ge -\inf(-A)$. On the other hand, for any $a \in A$, $-\inf(-A) \ge a$. Therefore $\sup(A) \le -\inf(-A)$. 
-\end{proof}
-
-
-\begin{definition}[Interval]
-\label{definition:ordered-vector-space-interval}
-    Let $(E, \le)$ be an ordered vector space and $x, y \in E$, then
-    \[
-    [x, y] = \bracs{z \in E| x \le z \le y}
-    \]
-
-    is the \textbf{order interval} with endpoints $x$ and $y$.
-\end{definition}
-
-\begin{definition}[Order Bounded]
-\label{definition:ordered-vector-space-bounded}
-    Let $(E, \le)$ be an ordered vector space and $A \subset E$, then $A$ is \textbf{order bounded} if there exists $x, y \in E$ such that $A \subset [x, y]$.
-\end{definition}
-
-\begin{definition}[Order Complete]
-\label{definition:order-vector-complete}
-    Let $(E, \le)$ be an ordered vector space, then $E$ is \textbf{order complete} if for any order bounded set $A \subset E$, $\sup (A)$ and $\inf (A)$ exist.
-\end{definition}
-
-\begin{definition}[Order Bounded Dual]
-\label{definition:order-bounded-dual}
-    Let $(E, \le)$ be an ordered vector space, then the space $E^b$ consisting of all linear functionals on $E$ that are bounded on order bounded sets is the \textbf{order bounded dual} of $E$.
-\end{definition}
-
-\begin{definition}[Order Dual]
-\label{definition:order-dual}
-    Let $(E, \le)$ be an ordered vector space and $\Phi^+ \in \hom(E; \real)$, then $\Phi^+$ is \textbf{positive} if for any $x \in E$ with $x \ge 0$, $\Phi^+(x) \ge 0$. The subspace $E^+ \subset \hom(E; \real)$ generated by the positive linear functionals on $E$ is the \textbf{order dual} of $E$. 
-\end{definition}
 
 
 \begin{definition}[Vector Lattice]
@@ -249,7 +192,7 @@
 \end{proof}
 
 
-\begin{proposition}[{{\cite[V.1.4]{SchaeferWolff}}}]
+\begin{proposition}
 \label{proposition:order-vector-dual}
     Let $(E, \le)$ be an ordered vector space. If for any $x, y \in E$ with $x, y \ge 0$, $[0, x] + [0, y] = [0, x + y]$, then:
     \begin{enumerate}
@@ -264,7 +207,7 @@
 
     
 \end{proposition}
-\begin{proof}
+\begin{proof}[Proof, {{\cite[V.1.4]{SchaeferWolff}}}. ]
     (1): Let $C = \bracs{x \in E|x \ge 0}$, $\Phi^+ \in E^b$, and
     \[
     \Phi^+: C \to [0, \infty) \quad x \mapsto \sup(f([0, x]))